Question 202130: Can anyone help me with ANY of these? Please...
1. Solve for the roots by factoring. Leave as an improper fraction.
2x^2 - x = 21 x = ___ and ___
2. Solve for the roots by completing the square.
3x^2 - 12x = 9 x = _______________
3. Solve for the roots by using the quadratic formula. (Leave as an improper fraction)
3x^2 + 11x - 4 = 0 x = ____ and _____
4. Solve for the roots by using the quadratic formula.
2(6-x) = x(x+5) x = ___________________
5. Use the given equation and associated roots to identify a, b, and c.
3x^2 + 11x - 4 = 0 and root1 = -4 , root2 = (1/3)
a= ____ b= ____ c= ____
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Standard form:
Start with: (x\ \ \ )) because that is the only way to get  as the high order term. Next we know that one of the signs has to be positive and the other negative so that you end up with a -21. But which is which? We know the factors of 21 are 7 and 3, and we also know that the lead coefficient, 2, has to be multiplied by one of those to get the middle term -- Ah ha! 2 times 3 is 6 plus -7 is -1, the value of the given 1st degree term coefficient, so:
(x + 3) = 0) .
Use the Zero Product Rule ( ), so:
 or
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Step 1, the lead coefficient must be 1, so divide through by the lead coefficient:
Step 2, put the constant term on the right hand side. This is already done for you, so go on to...
Step 3, divide the 1st degree coefficient by 2 and square the result: 4 divided by 2 is 2. 2 squared is 4. Add this result to both sides:
Step 4, factor the resulting perfect square trinomial on the left:
^2 = 7)
Step 5, take the square root of both sides remembering to consider both the positive and negative root:
Step 6, finish solving for 
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For all 2nd degree polynomial equations of the form  , the roots are given by:
Just do the substitutions:
 \pm sqrt{(11)^2 - 4(3)(-4)}}{2(3)} )
You can do your own arithmetic.
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 = x(x+5))
This one has to be put into standard form first.
Now just do the exact same thing we did in the previous problem except that now  ,  , and 
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The given roots are TMI; you don't need them. All you need is the equation  ,  is the lead coefficient, is the coefficient on the 1st degree term, and  is the constant term.
John
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